Subsets II @Leetcode

Subsets II 

//Given a collection of integers that might contain duplicates, nums, return all possible subsets.

//For example, If nums = [1,2,2], a solution is: [
/**
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

*/

Learn ideas from http://www.jiuzhang.com/solutions/subsets-ii/

1pay attention to how to solve the duplicates:
  if(pos != start && nums[pos] == nums[pos - 1])
          continue;
so when we do the for loop, and meet the element that is the start of the loop and it is equal to the former element, we just jump over this condition and continue the loop. Cause this condition has been solved when we do the first loop for the pos

2 also this is quite a unique dfs, it added the tem result for every dps function, cause when enter the loop, there must  be a unique condition

Here is the code. 

public class Solution {    public List<List<Integer>> subsetsWithDup(int[] nums) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        List<Integer> tem = new ArrayList<Integer>();        //get the length of the array        int length = nums.length;        //special case that the length of the array is zero        if(length == 0)          return result;        Arrays.sort(nums);        helper(0, tem, result, nums);        return result;    }    public void helper(int start,  List<Integer> tem, List<List<Integer>> result, int[] nums){      result.add(new ArrayList<Integer>(tem));      int length = nums.length;      for(int pos = start; pos < length; pos++){        if(pos != start && nums[pos] == nums[pos - 1])          continue;        tem.add(nums[pos]);        helper(pos + 1, tem, result, nums);        tem.remove(tem.size() - 1);      }    }}